Let $X$ be a connected, locally path connected space. I want to show that it is also path connected.

Following a suggestion in this answer, fix $a\in X$ and consider the set

$$U_a = \{x\in X : \text{there is a path from } x \text{ to } a\}$$

and show that it is clopen.

If the set is clopen, then we can decompose $X$ as a disjoint union of the open sets $U_a$ and $X\setminus U_a$. Since $X$ is connected and $U_a \not= \varnothing$ ($a\in U_a$), it follows that $U_a = X$.

My question concerns how to show that $U_a$ is clopen. I considered writing $U_a$ and $X\setminus U_a$ as unions of open sets as follows:

$$U_a = \bigcup \{\text{path-connected open neighbourhoods of $a$}\}$$

$$X\setminus U_a = \bigcup_{x\in X}\bigcup\{\text{path-connected open neighbourhoods of $x$ without $a$}\}$$

It is clear that any path-connected neighbourhood of $a$ is a subset of $U_a$, but if there is a path from $x$ to $a$, does it follow that $x$ is in a path connected neighbourhood of a?

  • 1
    $\begingroup$ I think it is easier to show that every element in $U_a$ has a neighbourhood that is contained in $U_a$, and the same thing for the complement. $\endgroup$
    – asdq
    Apr 30, 2019 at 17:09

1 Answer 1


Thanks to @asdq ’s suggestion, I can complete the proof as follows:

Any $x \in U_a$ has a path-connected neighborhood, say $V$. Any point in $V$ can be connected by a path to $a$ by first connecting it to $x$, and concatenating with the path to $a$. Hence, $V \subset U_a$, so $U_a$ is open.

If $x \notin U_a$, then there is a path-connected neighborhood $W$ of $x$ such that no point of $W$ connects to $a$ via a path; otherwise we could join that path to $x$, so $x \in U_a$. Therefore, $W \subset X\setminus U_a$, so $X\setminus U_a$ is open.


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